# Arithmetic progression

Arithmetic and Geometric progressions.

### Arithmetic progression

Find the sum of the numbers
1 + 2 + 3 + ... + 100 = ?

Math Tutor

Posts: 386
Joined: Sun Oct 09, 2005 11:37 am
Reputation: 14

Find $a_{1}$ and "d" in this arithmetic progression.
$|a_{5}:a_{3}=4$
$|a_{2}.a_{6}=-11$

charlie_eppes

Posts: 11
Joined: Sun Dec 14, 2008 9:05 pm
Reputation: 0

What is the value of $a_5$ and $a_2$,
$a_2$ and $a_6$,
using $a_1$ and $d$?

http://www.math10.com/en/algebra/arithm ... ssion.html

Math Tutor

Posts: 386
Joined: Sun Oct 09, 2005 11:37 am
Reputation: 14

$|{a_{5}}^{2}-{a_{3}}^{2}=72$
$|a_{5}+a_{8}=28$
Something is wrong in my answers. I can't define it.
And this
$|5a_{1}+10a_{5}=0$
$|S_{4}=14$

charlie_eppes

Posts: 11
Joined: Sun Dec 14, 2008 9:05 pm
Reputation: 0

[quote="charlie_eppes"]$|{a_{5}}^{2}-{a_{3}}^{2}=72$
$|a_{5}+a_{8}=28$

$|(a_{1}+4d)^{2}-(a_{1}+2d)^{2}=72$
$|a_{1}+4d+a_{1}+7d=28$

$2a_{1}+11d=28$
$a_{1}$ ?

${a_{1}}^{2}+8a_{1}d+16d^{2}-{a_{1}}^{2}+4a_{1}d-4d^{2}=72$
$12d^{2}+12a_{1}d=72$
$d^{2}+a_{1}d=6$
Then?

charlie_eppes

Posts: 11
Joined: Sun Dec 14, 2008 9:05 pm
Reputation: 0

From the first equation :
$a_1 = \frac{(28-11d)}{2}$
$a_1 = 14-5.5d$
then solve the quadratic equation and find d

Math Tutor

Posts: 386
Joined: Sun Oct 09, 2005 11:37 am
Reputation: 14

teacher wrote:From the first equation :
$a_1 = \frac{(28-11d)}{2}$
$a_1 = 14-5.5d$
then solve the quadratic equation and find d

Why $a_1 = 14-5.5d$
How to solve $d^2 +a_1 d=6$

charlie_eppes

Posts: 11
Joined: Sun Dec 14, 2008 9:05 pm
Reputation: 0

charlie_eppes wrote:
teacher wrote:From the first equation :
$a_1 = \frac{(28-11d)}{2}$
$a_1 = 14-5.5d$
then solve the quadratic equation and find d

10x!

charlie_eppes

Posts: 11
Joined: Sun Dec 14, 2008 9:05 pm
Reputation: 0

### sending solution

for this arthimetic progression,
sum=n(n+1)/2
sum=100(100+1)/2
sum=5050

parvez

Posts: 4
Joined: Sat May 30, 2009 8:23 am
Reputation: 1

a=1
d=1
n=100

s=100/2[2*1+(100-1)*1]
=50[2+99]
=50[101]
= 5050[/u]

shrinidhi

Posts: 2
Joined: Mon Aug 17, 2009 8:08 am
Reputation: 0

charlie_eppes wrote:Find $a_{1}$ and "d" in this arithmetic progression.
$|a_{5}:a_{3}=4$
$|a_{2}.a_{6}=-11$

$a_{2} = a_{1} + d; a_{3} = a_{1} +2d, a_{5} = a_{1} + 4d, a6 = a1 + 5d$

$a_{5}/a_{3} = (a_{1} + 4d)/(a_{1} + 2d) = 4 or 4a_{1} + 8d = a_{1} + 4d or 3a_{1} = -4d or a_{1} = -4d/3$
$(a_{1} + d) (a_{1} + 5d) = (-4d/3 + d)(-4d/3 + 5d) = (-d/3_)11d/3) = -(11d^2)/9 = -11 or d^2 = 8$
So d=3 and $a_{1} = -4$

MV RAO

Posts: 7
Joined: Wed Dec 09, 2009 5:01 am
Reputation: 0

charlie_eppes wrote:$|{a_{5}}^{2}-{a_{3}}^{2}=72$
$|a_{5}+a_{8}=28$
Something is wrong in my answers. I can't define it.
And this
$|5a_{1}+10a_{5}=0$
$|S_{4}=14$

a5^2 – a3^2 = 72, a5 + a8 = 28, find a1 and d
(a1 + 4d) + (a1 + 7d) = 2a1 + 11d = 28 or a1 = (28 – 11d)/2
(a1 + 4d)^2 – (a1 + 2d)^2 = a1^2 + 16d^2 + 8a1d – a1^2 – 4d^2 – 4a1d = 12d^2 + 4a1d = a2d^2 + 4d(28 – 11d)/2 = 12d^2 + 56d – 22d^2 = 72
Or -10d^2 + 56d = 72 or 5d^2 – 28d + 36 = (5d – 18 )(d – 2) = 0 or d = 2
a1 = (28 – 11d)/2 = 14 – 11 = 3

MV RAO

Posts: 7
Joined: Wed Dec 09, 2009 5:01 am
Reputation: 0

Hey Auster !!
Is it some kinda mathematical prank ?

Thanks
joetraff

Posts: 1
Joined: Fri Aug 13, 2010 10:04 am
Reputation: 0

### Re: Arithmetic progression

the sum of the first n term of arithmetic progression is given by Sn = n^2+3n
a) the first term
b) the second term
c)the common difference
how to solve this question??????

cik are u

### Re: Arithmetic progression

The problem seems to be incomplete.
Check up once again if the condition is correct, please.

Math Tutor

Posts: 386
Joined: Sun Oct 09, 2005 11:37 am
Reputation: 14

### Re: Arithmetic progression

Need help to solve this problem

Postby Auster1989 » Mon Aug 09, 2010 10:53 pm
hello,

My name is Auster from canada,Can you help me guys to solve this problem?

Think of a number between eight-hundred and ten and nine-hundred and sixty-three, then add one thousand and fifty-three.

sajid121

Posts: 10
Joined: Thu Apr 26, 2012 6:04 am
Reputation: 1

### Re: Arithmetic progression

Find the sum of the numbers
1 + 2 + 3 + ... + 100 = ?

1st method
First term(a)=1
Last term(b)=100
Numbers of terms(n)=100
Using formula
Sum=n/2(a+b)
=100/2(1+100)
=50*101
=5050

Alternative method
First term(a)=1
Common difference(d)=1
Numbers of terms(n)=100
Using formula
sum=n/2(2a+(n-1)d)
=100/2(2*1+(100-1)*1)
=50(2+99)
=50*101
=5050
Hence, ans=5050

sudarshan

Posts: 4
Joined: Sat Jun 30, 2012 7:50 am
Reputation: 1

### Re: Arithmetic progression

n(n+1)/2 = 100(100 + 1)/ 2 = 100 x 101 / 2 = 5050

Anne20

Posts: 1
Joined: Fri Aug 24, 2012 12:24 pm
Reputation: 0

### Re: Arithmetic progression

Let n= 100
so puttung value of n into n(n+1)/2

=> 100(100+1)/2

=> 100(101)/2

=> 10100/2

=> 5050

The sum of the given numbers is 5050.

leesajohnson

Posts: 218
Joined: Thu Dec 31, 2015 7:11 am
Location: London
Reputation: -33

### Re: Arithmetic progression

Given Arithmetic progression series had common difference as 1. hence it's sum till 100th term is
n(n+1)/2
= 100(100 + 1)/ 2
= 100 x 101 / 2
= 5050
http://www.techcrashcourse.com/2015/08/c-program-generate-arithmetic-series-sum.html

Guest

Next